# Probability

8th----

Bob's phone plan has a voicemail service with the storage capacity of maximum 2 voice messages.
Each morning, Bob checks and answers the voice messages according to his available time during that morning. The number of maximum messages that Bob can answer each morning is with probability po = 0.2, 1 with probability P1 = 0.3, 2 with probability P2 = 0.5. After Bob answers a message, the message is removed from the voice mailbox.
The number of messages that Bob receives the day before is O with probability qu = 0.1, 1 with probability qı = 0.5, and 2 with probability 42 = 0.4. If the voice mailbox is already full, all newly arrived messages will be discarded. For simplicity, assume that there is no voice messages received in the mornings during the time he checks.
A Markov chain model:
The above system can be modeled by a discrete-time homogeneous Markov chain as in the figure below:
11 P22 Poi P10 1 P12 P21 2 2 Po2 P20
In this Markov chain, we have three states V = {0, 1, 2} indicating the number of messages remaining in Bob's voicemail box after he checks his voicemails in the mornings. We derive that po = 0.7.
Question 1. What are the transition probabilities p01, p11, and p22 in this Markov chain?
P01 = ?
P11 = ?
P22 = ?
Notation: In all of the following questions, let X, denote the number of messages remaining in the mailbox after Bob checks and answers messages on day i, e.g. X2 = 1 indicates that I message is left in the mailbox after Bob checks on day 2
Question 2. What is the probability P(X2 = 1, X3 = 2, X4 = 0X1 = 0) (i.e. the probability that X, = 1, X3 = 2 AND X4 = 0 given that Xi = 0)?
P(X2 = 1, X3 = 2, X4 = 0|X1 = 0) = ?
Question 3. Given that X1 = 0, what is the probability that in each of the following three days there is never 2 messages remaining after he checks the mailbox, i.e. X: #2 for all a = 2, 3, 4?
Question 4. Given that X1 = 0, what is probability that X1 = 0? Note that there are three transitions in the process.

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1. 1.0.2,0.7,0.7
2.0.32
3.0.45
5.1/9

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3. 1) 0.22, 0.45, 0.45
2) 0.0242
3) ?
4)?

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5. Could someone please point out or provide a link to the other questions of the final exam?

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6. p01 = 0.22
p11 = 0.1
p22 = 0.08

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7. 2. 0.000512

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2. = 0.45
3. = .000512

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9. 4. = .343

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10. p01 = P(1 after day i+1 | 0 after day i). This can only happen if Bob receives one more message than he has time to answer, so p01 = q1*p0 + q2*p1 = 0.5*0.2 + 0.4*0.3 = 0.1 + 0.12 = 0.22.

p11 = P(1 after day i+1 | 1 after day i). Bob goes off to work on day i with one leftover message, and 24 hours later still only has one leftover message. If he didn't receive any messages, he must also not have had time to answer any (or his backlog would have disappeared). If he received any messages, then he must have answered one - note that if he received 2 messages, the last one would have bounced off the capacity limit, and he still would only have needed to answer one. So p11 = q0*p0 + (q1 + q2)*p1 = 0.1*0.2 + 0.9*0.3 = 0.02 + 0.27 = 0.29.

If Bob goes off to work on day i with 2 messages, then any more that come in will bounce off the capacity limit. If he answers any on day i+1, he will have less than 2. So p22 = p0 = 0.2.

When finding the other probabilities involved, it's important to note the overflow cases. Bob can't answer more messages than he actually receives, so any morning when he can answer 2 must result in 0 messages. Any messages over the capacity limit of 2 are lost, so the motion from state 2 is entirely determined by how many he answers - i.e., p2j = p(2-j).

2. The probability of going S0-S1-S2-S0, given that you start in S0, is entirely described by the product of transition properties p01*p12*p20.

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11. 3) 0.7367
p00^3+p00^2*p01+p00*p01*p11+p00*p01*p10+p01*p11^2+p01*p10*p00+p01*p10*p01
4) 0.6239

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